A parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E _2$. The ratio $E _2 / E _1$ is
$2: 1$
$1: 2$
$1: 4$
$2: 3$
Two capacitors each of $1\,\mu F$ capacitance are connected in parallel and are then charged by $200\;volts$ $d.c.$ supply. The total energy of their charges (in $joules$) is
$(a)$ A $900 \;p\,F$ capacitor is charged by $100 \;V$ battery [Figure $(a)$]. How much electrostatic energy is stored by the capacitor?
$(b)$ The capacitor is disconnected from the battery and connected to another $90\; p\,F$ capacitor [Figure $(b)$]. What is the electrostatic energy stored by the system?
The energy stored in the electric field produced by a metal sphere is $4.5\, J$. lf the sphere contains $4\,\mu C$ charge, its radius will be.......$mm$ : [Take : $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}\,N - {m^2}\,/{C^2}\, ]$
The plates of a parallel plate capacitor have an area of $90 \,cm ^{2}$ each and are separated by $2.5\; mm .$ The capacitor is charged by connecting it to a $400\; V$ supply.
$(a)$ How much electrostatic energy is stored by the capacitor?
$(b)$ View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.
A parallel plate capacitor has a uniform electric field ' $\overrightarrow{\mathrm{E}}$ ' in the space between the plates. If the distance between the plates is ' $\mathrm{d}$ ' and the area of each plate is ' $A$ ', the energy stored in the capacitor is : $\left(\varepsilon_{0}=\right.$ permittivity of free space)